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For on-line tutorial
help with kinetics, there are four modules available in the Kinetics
section of chemcal and in particular, the module
Kinetics and Reaction Mechanisms.
Shortcut to Questions
Q: 1 2 3
reaction represented by 2A + B → C it is found that doubling
the concentrations of both A and B quadruples the rate of reaction
at a fixed temperature. Doubling the concentration of B alone does
not affect the rate of reaction.
Write an expression for the rate law of the reaction
Suggest a sequence of steps compatible with this behaviour.
A probable mechanism for the reaction -
NO2(g) + CO(g) →
CO2(g) + NO(g)
for temperatures below about 773 K involves the following steps -
NO3 + NO (slow) .....(1)
NO3 + CO →
NO2 + CO2 (fast).....(2)
(a) Write the rate
equation for the reaction if this mechanism is correct.
(b) Give the order
of the reaction below 773 K with respect to
(c) At temperatures
well above 773 K the probable mechanism for the complete reaction
consists of only one step. Write the rate equation for the reaction
if this mechanism is correct.
(d) Give the order
of the reaction in (c) with respect
At high temperatures, nitrogen oxide and hydrogen react according to the stoichiometric equation -
2NO(g) + 2H2(g) →
N2(g) + 2H2O(g)
A likely mechanism for the reaction involves the following series of steps -
NO(g) + NO(g) N2O2(g) fast, reversible (1)
N2O2(g) + H2(g) →
N2O(g) + H2O(g) slow (2)
N2O(g) + H2(g) →
N2(g) +H2O(g) fast (3)
What is the rate equation for the reaction if the above mechanism is correct?
A certain reaction has the following rate equation -
rate = k [H2][Cu2+]2
(a) What is the overall
order of the reaction?
(b) What would be the effect on
the rate of reaction if-
(i) The concentration
of hydrogen were tripled, all other concentrations being held constant?
(ii) The concentrations of all
three species in the rate equation were simultaneously doubled?
The rate of the reaction H2O2 + 2I- + 2H+ →
I2 + 2H2O can be determined by using spectrophotometric means to measure the time taken, after mixing reactants, for [I2] to reach 1.00 x 10-5 M. It can be assumed (for suitable starting concentrations) that the rate is very nearly constant for this period.
(a) In an experiment
in which [H2O2] = 0.010 M, [I-]
= 0.010 M, and [H+] = 0.100 M, the time taken for [I2]
to reach 1.00 x 10-5 M was 5.7 second. Calculate the
rate of reaction (d[I2]/dt).
(b) When the experiment
was repeated with [H2O2] = 0.0050 M, [I-]
= 0.0100 M and [H+] = 0.100 M, the time was 11.5 second.
Calculate the rate now.
(c) From these calculations show
that the reaction rate depends on [H2O2] raised
to the first power, i.e. that R ∝
(d) Given the further
information that the rate law is R = k [H2O2]
[H+][I-], calculate the rate constant, k.
(e) Predict the rate of reaction if
[H2O2] = 0.0500 M, [H+] = 0.100
M and [I-] = 0.0200 M.
The thermal decomposition of nitrogen dioxide can be represented
by the equation
2NO(g) + O2(g)
From an experimental study, the following results were obtained for the rate
of the reaction at 700 K.
Initial rate of
(a) Use this data to deduce a general
expression for the rate of the reaction.
(b) What is the numerical value of
the rate constant at 700 K?
For the reaction RCl + OH- →
ROH + Cl- the following rate equations were determined experimentally for dilute solutions.
R = (CH3)3C-
R = CH3-
Rate = k1.[RCl]
Rate = k2.[RCl]
(k1 and k2 are rate constants)
For each reaction indicate by writing "yes" or "no" in
each of columns whether the following possible reaction mechanisms
are compatible with the rate equation.
R is (CH3)3C-
R is CH3-
+ Cl- (slow)
+ OH- → ROH (fast)
+ Cl- (fast)
+ OH- → ROH (slow)
RCl + OH-
→ ROH + Cl-
to Kinetics (Advanced) Questions
rate law can only be determined by experiments such as those given
in the data, and not from the coefficients in the stoichiometric
In the experiment where
the concentration of A (written as [A]) is not altered but the
[B] is doubled, the result is no change in the rate of reaction.
Hence the rate is not determined by [B]. i.e. the rate is zero
order with respect to B.
In the experiment where
both [A] and [B] are doubled, the rate increases by a factor of
4. However, as the rate is not affected by changes in [B], this
increase in rate must be attributable entirely to the changed
[A]. Thus doubling [A] causes a fourfold increase in rate, so
the rate ∝ [A]2 or
the reaction is second order with respect to A.
Thus the rate law is
rate ∝ [A]2[B]0
or rate = k[A]2.
The reaction being second
order with respect to A means that in the rate determining step, two
molecules of reactant A must combine to form an intermediate which
could be represented as A.A and as this is the rate-determining step,
it must be the slow step. In a subsequent step, one of these A.A intermediate
species reacts with one molecule of B (from the stoichiometric equation)
to form one molecule of the product, C. This second step must be fast
as the rate is not affected by [B]. This mechanism could be shown
by the following sequence of steps:
A + A → A.A (slow
A.A + B → C (fast
In the slow (rate-determining) step (1), two molecules of NO2
combine to form the intermediate species, NO3 plus a
molecule of one product, NO. The reaction is therefore second order
with respect to NO2 or rate = k[NO2]2.
second order see (a) above
(ii) As CO does not feature as a reactant
in the slow step but only appears in the fast step, the order of
the reaction with respect to CO is zero.
(c) If the mechanism at the higher temperature is
just a single step, then it must be the same as shown in the stoichiometric
equation and so
rate ∝ [NO2][CO] or
rate = k[NO2][CO].
(d) From the rate equation in part (c),
the reaction is now first order with respect to both NO2
From the slow step, rate
The rate equation in
terms of these species would be rate = k2[N2O2][H2].
However, one of these species, N2O2, is an
intermediate produced by the equilibrium step 1 and not one of the
original reactants whose concentrations can be measured before the
can be obtained in terms of the concentrations of the initial reactants
using the equilibrium constant for step 1 as follows:
Kc = [N2O2]
= Kc x [NO]2
Substituting this in
the rate equation rate = k2 [N2O2][H2]
Rate = k2Kc[NO]2[H2]
or, combining the constants,
Rate = k[NO]2[H2].
This mechanism is an
illustration of how a reaction can be third order overall while
the rate determining step is only second order. In reactions, it
is highly unlikely that more than two species will collide simultaneously
with sufficient energy and with the required impact orientation
for them to react, so a third order mechanistic step is highly improbable.
So while this reaction mechanism would be third order overall, it
is only second order in the slow step, (2).
overall order for the reaction is obtained by adding all the exponents
of the concentrations in the rate equation.
The rate equation here
is rate = k[H2][Cu2+]2[H+]-1,
so overall order = 1
+ 2 +(-1) = 2, i.e. second order.
This is an example of
a reaction where increasing the concentration of a reactant (the
H+) causes the reaction to proceed more slowly as seen
by its negative exponent in the rate equation. Thus the greater
the concentration of H+, the more the reaction is inhibited.
Increasing the [H2] by a factor of three would increase
the rate by a factor of 3 as the reaction is first order with respect
Doubling [H2] doubles the rate as the reaction is first
order with respect to H2. Doubling [Cu2+]
increases the rate by a further factor of four as the reaction is
second order with respect to Cu2+. Doubling [H+]
will halve the rate as the order is -1 with respect to H+.
Combining these changes, the overall change in rate is by a factor
of 2 x 4 x ½ = four times the original rate.
The change in [I2] during the time interval of 5.7 second
= final concentration
- initial concentration
= 1.00 x 10-5
- 0 M = 1.00 x 10-5 M.
∴ rate of formation
of I2 = 1.00 x 10-5 / 5.7 M s-1
= 1.8 x 10-6 M s-1.
The time interval is now 11.5 seconds to produce the same [I2].
∴ Rate = 1.00 x 10-5
/ 11.5 = 8.7 x 10-7 M s-1.
In the two experiments, only [H2O2] was altered.
In the second experiment, halving its concentration as used in the
first experiment caused the rate to halve, so the reaction must be
directly proportional to [H2O2], or
Rate ∝ [H2O2]1.
that the rate law is rate = k[H2O2][H+][I-],
k can be calculated by substituting either of the sets of
data given in (a) or (b) above.
Using data from (b),
8.7 x 10-7 = k (0.0050) (0.100) (0.010) = k
x 5.0 x 10-6
k = 8.7 x 10-7 / 5.0 x 10-6 = 1.7 x
10-1 M-2 s-1.
this value of k in the rate equation,
Rate = 1.8 x 10-1
(0.0500) (0.100) (0.0200)
= 1.8 x 10-5
Alternatively, one could
use the method of direct comparison of this new data with that in
(b) above and note that only [I-] is changed,
being increased from 0.0100 M to 0.0200 M. Given the reaction is
first order with respect to I-, this doubling of [I-]
would double the rate calculated in (b) so
rate = 2 x 8.7 x 10-7
= 1.7 x 10-5 M s-1.
In this type of problem, check to see if the initial concentration
data for the series of experiments has been changed by simple factors
such as doubling. In this situation, the rate equation can be deduced
merely by inspection.
However, the data given
in this problem does not have a simple relationship between the
initial concentrations used in the two experiments, so the deduction
of the rate law is a little more complicated. Using the rate law,
rate = k[NO2]n, substitution of the
two sets of data results in two equations with two unknowns, k
and n, which can therefore be obtained.
From the first experiment,
0.0031 = k (0.0225)n ................(1)
From the second experiment,
0.0016 = k (0.0162)n ..........(2)
Divide equation (1)
by equation (2)
0.0031 / 0.0016 =
k (0.0225)n /k
1.94 = (1.39)n
Take logs of both sides:
log (1.94) = n log(1.39)
0.286 = n x 0.143
∴ n = 2.0
and so the rate law is
rate = k[NO2]2.
To evaluate k,
substitute either set of data in the rate law. Using data from
the first experiment,
0.0031= k (0.0225)2
k = 6.1 M-1 s-1.
When R is (CH3)3C-,
the rate is directly proportional to [RCl], so only RCl should appear
in the slow step of the mechanism equations for this reaction. When
R is CH3-, the rate is proportional to both [RCl] and
[OH-], so both RCl and OH- must appear in
the slow step of any proposed mechanism for this reaction. Three
proposed mechanisms are given.
R+ + Cl- (slow)
R+ + OH-
→ ROH (fast)
This mechanism is consistent
with R being (CH3)3C- as only RCl appears
in the slow step but not consistent with R being CH3-.
RCl → R+ + Cl-
R+ + OH-
→ ROH (slow)
In this proposed mechanism,
the slow (rate-determining) step includes R+ which
is not one of the initial reactants, so its concentration must
be expressed in terms of [RCl] using the equilibrium constant
for the first (fast) step as follows:
= [R+][Cl-]/[RCl], so [R+] =
The rate equation from
the second step is rate = k[R+][OH-]
Rate = kKc[RCl][OH-]/[Cl-].
This rate equation
requires the rate to be directly proportional to both [RCl] and
[OH-] and inversely proportional to [Cl-].
(i.e. the presence of Cl- would inhibit the reaction.).
This is not consistent with the observed rate data for either
reaction and therefore is not a possible mechanism in either case.
+ OH- → ROH + Cl-
If the reaction mechanism
consists of just one step, then this must be the rate-determining
step. Here the rate equation would be
Rate = k[RCl][OH-]
which is not consistent with the data when R is (CH3)3C-
but would be consistent with the data for R = CH3-.